**Question 6:**In the diagram below,

*CD*is an arc of a circle with centre

*O*.

Determine the area of the shaded region.

$\left(\text{Use}\pi =\frac{22}{7}\right)$

*Solution*:$\begin{array}{l}\text{Areaofsector}=\text{Areaofcircle}\times \frac{{72}^{o}}{{360}^{o}}\\ \text{}=\frac{22}{7}\times {\left(10\right)}^{2}\times \frac{{72}^{o}}{{360}^{o}}\\ \text{}=\frac{440}{7}{\text{cm}}^{2}\\ \text{Areaof}\Delta OBD=\frac{1}{2}\times 6\times 8\\ \text{}=24{\text{cm}}^{2}\\ \text{Areaofshadedregion}=\frac{440}{7}-24\\ \text{}=38\frac{6}{7}{\text{cm}}^{2}\end{array}$

**Question 7:**In diagram below,

*ABC*is a semicircle with centre

*O*.

Calculate the area, in cm

^{2}, of the shaded region.

$\left(\text{Use}\pi =\frac{22}{7}\right)$

*Solution*:$\begin{array}{l}\angle ACB={90}^{o}\\ AB=\sqrt{{6}^{2}+{8}^{2}}\\ \text{}=\sqrt{100}\\ \text{}=10\text{cm}\\ \text{Radius}=10\xf72\\ \text{}=5\text{cm}\\ \\ \text{Theshadedregion}\\ =\left(\frac{1}{2}\times \frac{22}{7}\times 5\times 5\right)-\left(\frac{1}{2}\times 6\times 8\right)\\ =39\frac{2}{7}-24\\ =15\frac{2}{7}{\text{cm}}^{2}\end{array}$

**Question 8:**In diagram below,

*ABC*is an arc of a circle centre

*O*

The radius of the circle is 14 cm and

*AD*= 2

*DE*.

Calculate the perimeter, in cm, of the whole diagram.

$\left(\text{Use}\pi =\frac{22}{7}\right)$

*Solution*:$\begin{array}{l}\text{Lengthofarc}ABC\\ =\frac{3}{4}\times 2\pi r\\ =\frac{3}{4}\times 2\times \frac{22}{7}\times 14\\ =66\text{cm}\\ \\ \text{Perimeterofthewholediagram}\\ =16+8+8+66\\ =98\text{cm}\end{array}$

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**Question 9:**

In diagram below,

*KLMN*is a square and

*KLON*is a quadrant of a circle with centre

*K*.

Calculate the area, in cm

^{2}, of the coloured region.

$\left(\text{Use}\pi =\frac{22}{7}\right)$

*Solution*:$\begin{array}{l}\text{Areaofthecolouredregion}\\ =\frac{{45}^{o}}{{360}^{o}}\times \pi {r}^{2}\\ =\frac{{45}^{o}}{{360}^{o}}\times \frac{22}{7}\times {14}^{2}\\ =77{\text{cm}}^{\text{2}}\end{array}$

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**Question 10:**

Diagram below shows two quadrants,

*AOC*and

*EOD*with centre

*O*.

Sector

*AOB*and sector

*BOC*have the same area.

Calculate the area, in cm

^{2}, of the coloured region.

$\left(\text{Use}\pi =\frac{22}{7}\right)$

*Solution*:$\begin{array}{l}\text{Areaofthesector}AOB=\text{Areaofthesector}BOC\\ \text{Therefore,}\angle AOB=\angle BOC\\ \text{}={90}^{o}\xf72\\ \text{}={45}^{o}\\ \text{Areaofthecolouredregion}\\ =\frac{{45}^{o}}{{360}^{o}}\times \frac{22}{7}\times {16}^{2}\\ =100\frac{4}{7}{\text{cm}}^{2}\end{array}$